Brian Blais wrote: > Hello, > > In my attempt to learn python, migrating from matlab, I have the following > problem. > Here is what I want to do, (with the wrong syntax): > > from numpy import * > > t=arange(0,20,.1) > x=zeros(len(t),'f') > > idx=(t>5) > tau=5 > x[idx]=exp(-t[idx]/tau) # <---this line is wrong (gives a TypeError) > > #------------------ > > what is the best way to replace the wrong line with something that works: > replace all > of the values of x at the indices idx with exp(-t/tau) for values of t at > indices idx? > > I do this all the time in matlab scripts, but I don't know that the pythonic > preferred method is. > Brian, First, let me warn you I only know the old Numeric, and I haven't made the jump to Numpy yet. I'm told the syntax is about the same though. Using nonzero, put, and take, I can do what you want like this:
import Numeric as num t=num.arange(0,20,.1) x=num.zeros(len(t),'f') idx=num.nonzero(t>5) tau=5. num.put(x,idx,num.exp(-num.take(t,idx)/tau)) Kinda messy compared to Matlab. Do you know about Octave (www.octave.org)? It tries to be compatible with Matlab syntax. Of course it lacks the other niceties of python. Paul Probert -- http://mail.python.org/mailman/listinfo/python-list
